<u>Given:</u>
Concentration of HNO3 = 7.50 M
% dissociation of HNO3 = 33%
<u>To determine:</u>
The Ka of HNO3
<u>Explanation:</u>
Based on the given data
[H+] = [NO3-] = 33%[HNO3] = 0.33*7.50 = 2.48 M
The dissociation equilibrium is-
HNO3 ↔ H+ + NO3-
I 7.50 0 0
C -2.48 +2.48 +2.48
E 5.02 2.48 2.48
Ka = [H+][NO3-]/HNO3 = (2.48)²/5.02 = 1.23
Ans: Ka for HNO3 = 1.23
Answer:
ruthenium (Ru), rhodium (Rh), palladium (Pd), osmium (Os), iridium (Ir), platinum (Pt), gold (Au), silver (Ag).
Explanation:
Answer:
2-ethoxy-2-methylpropan-1-ol
Explanation:
On this reaction, we have an "<u>epoxide"</u> (2-methyl-1,2-epoxypropane). Additionally, we have <u>acid medium</u> (due to the sulfuric acid
). The acid medium will produce the <u>hydronium ion</u> (
). This ion would be attacked by the oxygen of the epoxide. Then a <u>carbocation</u> would be produced, in this case, the most stable carbocation is the <u>tertiary one</u>. Then an <u>ethanol</u> molecule acts as a nucleophile and will attack the carbocation. Finally, a <u>deprotonation </u>step takes place to produce <u>2-ethoxy-2-methylpropan-1-ol</u>.
See figure 1
I hope it helps!
Explanation:
Volume is the quantity of three-dimensional